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December 4, 2012

RMO 2012 solution to Question No. 6

6. Find all positive integers n such that (3^{2n} + 3 n^2 + 7 ) is a perfect square.
We use the fact that between square of two consecutive numbers there exist no perfect square. That is between (k^2 ) and ((k+1)^2 ) there is no square.
Note that (3^{2n} = (9^n)^2 ) and  ((9^n + 1)^2 ) are two consecutive perfect square and (3^{2n} + 3 n^2 + 7 ) is always a number between them for n > 2 (easily proved by induction).

Hence the only solution is n = 2.

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